A difference of 2.3 eV separates two energy levels in an atom

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atoms

#1

A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?


#2

Separation of two energy levels in an atom,∆E = 2.3 eV = 2.3 x 1.6 x ${{10}^{-19}}$ = 3.68 x ${{10}^{-19}}$ J
Let V be the frequency of radiation emitted, when the atom transits from the upper level to the lower level.
We have the relation for energy as, ∆E = h v
where, h = 6.63 x ${{10}^{-34}}$ J-s
v = ∆E / h = 3.68 x ${{10}^{-19}}$ / 6.63 x ${{10}^{-34}}$ = 5.6 x ${{10}^{14}}$ Hz
Hence, the frequency of the radiation is 5.6 x ${{10}^{14}}$ Hz